Question: Simplify and expand the following expression: $ \dfrac{5}{t - 10}- \dfrac{5}{3t + 27}+ \dfrac{t}{t^2 - t - 90} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{5}{3t + 27} = \dfrac{5}{3(t + 9)}$ We can factor the quadratic in the third term: $ \dfrac{t}{t^2 - t - 90} = \dfrac{t}{(t - 10)(t + 9)}$ Now we have: $ \dfrac{5}{t - 10}- \dfrac{5}{3(t + 9)}+ \dfrac{t}{(t - 10)(t + 9)} $ The least common multiple of the denominators is: $ (t - 10)(t + 9)$ In order to get the first term over $(t - 10)(t + 9)$ , multiply by $\dfrac{3(t + 9)}{3(t + 9)}$ $ \dfrac{5}{t - 10} \times \dfrac{3(t + 9)}{3(t + 9)} = \dfrac{15(t + 9)}{(t - 10)(t + 9)} $ In order to get the second term over $(t - 10)(t + 9)$ , multiply by $\dfrac{t - 10}{t - 10}$ $ \dfrac{5}{3(t + 9)} \times \dfrac{t - 10}{t - 10} = \dfrac{5(t - 10)}{(t - 10)(t + 9)} $ In order to get the third term over $(t - 10)(t + 9)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{t}{(t - 10)(t + 9)} \times \dfrac{3}{3} = \dfrac{3t}{(t - 10)(t + 9)} $ Now we have: $ \dfrac{15(t + 9)}{(t - 10)(t + 9)} - \dfrac{5(t - 10)}{(t - 10)(t + 9)} + \dfrac{3t}{(t - 10)(t + 9)} $ $ = \dfrac{ 15(t + 9) - 5(t - 10) + 3t} {(t - 10)(t + 9)} $ Expand: $ = \dfrac{15t + 135 - 5t + 50 + 3t}{3t^2 - 3t - 270} $ $ = \dfrac{13t + 185}{3t^2 - 3t - 270}$